How to calculate this angular velocity problem

AI Thread Summary
The discussion revolves around calculating the angular velocity of a turntable after two blocks stick to it. The initial calculations for angular momentum and moment of inertia were incorrect, as the contributions of the blocks at the outer radius were not properly accounted for. The correct formula for the moment of inertia of a mass at a distance from the center is I = MR^2, which applies to the blocks as well. After realizing the need to combine the moments of inertia of the turntable and the blocks, the user found the correct approach to solve for the new angular velocity. The conversation highlights the importance of accurately calculating the moment of inertia in rotational dynamics problems.
karzy
Messages
4
Reaction score
0
A 2.3kg , 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500g blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick.
What is the turntable's angular velocity, in rpm, just after this event?I first cacluted the angular momentum as .120428 using I*W so I then set that equal to I*W where I=.5(3.3)(.1)^2
and w is 2pi(frequency) and tried solving for frequency. and converting that back to rpm. But 69.7 rpm wasn't correct so I'm missing something or approaching this wrong!

Any suggestions?
 
Last edited by a moderator:
Physics news on Phys.org
What's the new moment of inertial after the blocks stick to the turntable?

Chet
 
i used the formula (1/2)(M)R^2 which I used 3.3 for the mass and .1 for the R which gave me .0165.
 
karzy said:
i used the formula (1/2)(M)R^2 which I used 3.3 for the mass and .1 for the R which gave me .0165.
This formula applies to the disk, but not to the blocks. The blocks are at the outside radius, so their contributions to the moment of inertial (per unit mass) are greater. Do you know the formula for the moment of inertial of a mass at a distance r from the center of rotation?

Chet
 
I=MR^2 where m is the mass and r is distance from the center?
 
Ahhh i just got it! I needed to add the three Ineritas together then solve for the velocity.Thanks for the help!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top